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Question: Prove that the expression \(a\left( \sin B-\sin C \right)+b\left( \sin C-\sin A \right)+c\left( \sin...

Prove that the expression a(sin⁡B−sin⁡C)+b(sin⁡C−sin⁡A)+c(sin⁡A−sin⁡B)=0a\left( \sin B-\sin C \right)+b\left( \sin C-\sin A \right)+c\left( \sin A-\sin B \right)=0 is true for any ΔABC\Delta ABC

Explanation

Solution

It is given ABC is a triangle. So, we can use the properties of triangle and say that the sum of all interior angles is180∘{{180}^{\circ }} .We can also use the sine rule of triangle
sin⁡Aa=sin⁡Bb=sin⁡Cc=k\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=k
By basic knowledge of algebra we can convert the above relation into relation between sin and its respective side as:
sin (A) = ak
By basic knowledge of algebra we can convert the above relation into relation between sin and its respective side as:
sin (b) = bk
By basic knowledge of algebra we can convert the above relation into relation between sin and its respective side as:
sin (c) = ck
Use the formulae given above to solve the question in a simpler manner.

Complete step-by-step answer :
Given condition:
ABC is a triangle
As it is a triangle, we can use the sum of angles as 180 degrees.
By basic knowledge of properties of triangles, we know that:
Sum of all interior angles must be equal to 180∘{{180}^{\circ }}
Now by subtracting C on both sides of equation we get:
A+B=π−CA+B=\pi -C
Now by subtracting A on both sides of equation (i) we get:
B+C=π−AB+C=\pi -A
Now by subtracting B on both sides of equation (i) we get:
A+C=π−BA+C=\pi -B
The value π\pi in degrees is equal to 180 degrees. By basic operations like subtraction we derived 3 equations. Now we can use these to solve further and then make the required expression simpler and reach our result faster.
By basic knowledge of properties of triangle, we know sine rule
sin⁡Aa=sin⁡Bb=sin⁡Cc=k\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=k
By cross multiplying each term we get three equations, they are:
sinA⁡=ak sinB=bk sinC=ck \begin{aligned} & \operatorname{sinA}=ak \\\ & sinB=bk \\\ & sinC=ck \\\ \end{aligned}
Required expression which is given question, we have to prove:
a(sin⁡B−sin⁡C)+b(sin⁡C−sin⁡A)+c(sin⁡A−sin⁡B)=0a\left( \sin B-\sin C \right)+b\left( \sin C-\sin A \right)+c\left( \sin A-\sin B \right)=0
By taking left hand side separately to prove it to be 0:
a(sin⁡B−sin⁡C)+b(sin⁡C−sin⁡A)+c(sin⁡A−sin⁡B)a\left( \sin B-\sin C \right)+b\left( \sin C-\sin A \right)+c\left( \sin A-\sin B \right)
By substituting 3 equations which are derived using the sine rule we get:
a(bk−ck)+b(ck−ak)+c(ak−bk)a\left( bk-ck \right)+b\left( ck-ak \right)+c\left( ak-bk \right)
By simplifying each term in the above expression, we get that:
abk−ack+bck−bak+cak−cbkabk-ack+bck-bak+cak-cbk
By grouping each term of same coefficient together, we get:
abk−abk+bck−bck+ack−ackabk-abk+bck-bck+ack-ack
By taking “k” common and grouping two terms in one bracket each:
k(ab−ab)+k(bc−bc)+k(ac−ac)k\left( ab-ab \right)+k\left( bc-bc \right)+k\left( ac-ac \right)
By simplifying the above equation, we get that this is
k(0)+k(0)+k(0)k\left( 0 \right)+k\left( 0 \right)+k\left( 0 \right)
By adding all terms in the above equation, we can say’0’.
By equating this to the left-hand side term, we get:
a(sin⁡B−sin⁡C)+b(sin⁡C−sin⁡A)+c(sin⁡A−sin⁡B)=0a\left( \sin B-\sin C \right)+b\left( \sin C-\sin A \right)+c\left( \sin A-\sin B \right)=0
Hence, we proved the required equation given in question.

Note :Use the sine rule carefully as the proportionality constant k is same because they all are equal. The relations derived from sine rule are the basic equations used here in this solution. So if you make a mistake in one equation you may lose the result you get. So carefully derive relations between sin and its corresponding angle.